What is inverse oflaplace (ce^{-as})/(s^2) ?

The answer to inverse oflaplace (ce^{-as})/(s^2) is \H(t-a)c(t-a)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (ce^{-as})/(s^2)

inverselaplace

H (t - a) c (t - a)

Solution steps

L^{- 1} {\frac{c e ^{- a s}}{s ^{2}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - a) L^{- 1} {\frac{c}{s ^{2}}} (t - a)

L^{- 1} {\frac{c}{s ^{2}}} : c t

= H (t - a) c (t - a)

What is inverse oflaplace (ce^{-as})/(s^2) ?
The answer to inverse oflaplace (ce^{-as})/(s^2) is \H(t-a)c(t-a)